On a train moving with acceleration `10 m//s^(2)`, a ball starts rolling on floor of train along the width of the train with angular acceleration `2 rad//s^(2)`, radius of ball is `1m`. Find the acceleration of the top point of ball at the time `t=3s` as seen from ground

Options

A. `10m//s^(2)`
B. `10sqrt(14)m//s^(2)`
C. `100m//s^(2)`
D. `sqrt(1396)m//s^(2)`

Answer

B

Explanation

Net acceleration of ball is :

`vec(a)`ball`//`ground =`vec(a)` ball`//``CM+vec(a)CM//` ground `vec(a)` ball `//CM=underset("tan gential acceleration")(vec(a)t)+underset("centripetal acceleration")(vec(a)C//CM)`
`|vec(a)"ball/ground"|=sqrt(a_(CM)^(2)+(Ralpha)^(2)+(omega^(2)R)^(2))`
`=sqrt(10^(2)+(1xx2)^(2)+[(2xx3)^(2)xx1^(2)]^(2))`
`=sqrt(1400) m//s^(2)`